Given below are two statements :
Statement I : Mohr's salt is composed of only three types of ions-ferrous, ammonium and sulphate.
Statement II : If the molar conductance at infinite dilution of ferrous, ammonium and sulphate ions are $$x_1$$, $$x_2$$ and $$x_3$$ S cm$$^2$$ mol$$^{-1}$$, respectively then the molar conductance for Mohr's salt solution at infinite dilution would be given by $$x_1 + x_2 + 2x_3$$
In the light of the given statements, choose the correct answer from the options given below :

Mohr’s salt is $$FeSO_4\cdot (NH_4)_2SO_4\cdot 6H_2O$$.

When it dissolves in water, the crystal breaks into ions according to the stoichiometric formula:
$$FeSO_4\cdot (NH_4)_2SO_4 \rightarrow Fe^{2+} + 2\,NH_4^{+} + 2\,SO_4^{2-}$$

Thus only three kinds of ions are present—ferrous ($$Fe^{2+}$$), ammonium ($$NH_4^{+}$$) and sulphate ($$SO_4^{2-}$$).
So, Statement I is correct.

At infinite dilution, Kohlrausch’s Law states that the molar conductance of an electrolyte equals the sum of the ionic conductances, each multiplied by the number of that ion present in one formula unit:
$$\Lambda_m^{\infty}= \sum n_i\,\lambda_i^{\infty}$$

For Mohr’s salt the stoichiometric coefficients are 1 for $$Fe^{2+}$$, 2 for $$NH_4^{+}$$ and 2 for $$SO_4^{2-}$$. If their individual molar ionic conductances are $$x_1$$, $$x_2$$ and $$x_3$$ S cm$$^2$$ mol$$^{-1}$$, then
$$\Lambda_m^{\infty}= 1\,(x_1) + 2\,(x_2) + 2\,(x_3)= x_1 + 2x_2 + 2x_3$$

Statement II proposes $$x_1 + x_2 + 2x_3$$, which neglects the second ammonium ion. Hence Statement II is false.

Therefore, Statement I is true while Statement II is false, which matches Option C.